2.1: Fibonacci's Rabbits
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In 1202, Fibonacci proposed the following puzzle, which we paraphrase here:
A man put a male-female pair of newly born rabbits in a field. Rabbits take a month to mature before mating. One month after mating, females give birth to one male-female pair and then mate again. No rabbits die. How many rabbit pairs are there after one year?
The growth of Fibonacci’s rabbit population is presented in Table 2.1. At the start of each month, the number of juvenile, adult, and total number of rabbits are shown. At the start of January, one pair of juvenile rabbits is introduced into the population. At the start of February, this pair of rabbits have matured and mate. At the start of March, this original pair of rabbits give birth to a new pair of juvenile rabbits. And so on.
If we let Fn be the total number of rabbit pairs at the start of the nth month, then the number of rabbits pairs at the start of the 13 th month will be the solution to Fibonacci’s puzzle. Examining the total number of rabbit pairs in Table 2.1, it is evident that
Fn+1=Fn+Fn−1.
This second-order linear difference equation requires two initial conditions, which are given by F1=F2=1. The first thirteen Fibonacci numbers, read from the table,
month | J | F | M | A | M | J | J | A | S | O | N | D | J |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
juvenile | 1 | 0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 |
adult | 0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 | 144 |
total | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 | 144 | 233 |
are given by
1,1,2,3,5,8,13,21,34,55,89,144,233,…
where F13=233 is the solution to Fibonacci’s puzzle.
Let us solve (2.1.1) for all the F′n s. To solve this equation, we look for a solution of the form Fn=λn. Substitution into (2.1.1) yields
λn+1=λn+λn−1
or after division by λn−1 :
λ2−λ−1=0
with solution
λ±=1±√52
Define
Φ=1+√52=1.61803…
and
ϕ=√5−12=Φ−1=0.61803…
Then λ+=Φ and λ−=−ϕ. Also, notice that since Φ2−Φ−1=0, division by Φ yields 1/Φ=Φ−1, so that
ϕ=1Φ
As in the solution of linear homogeneous differential equations, the two values of λ can be used to construct a general solution to the linear difference equation using the principle of linear superposition:
Fn=c1Φn+c2(−ϕ)n.
Extending the Fibonacci sequence to F0=0 (since F0=F2−F1 ), we satisfy the conditions F0=0 and F1=1 :
c1+c2=0,c1Φ−c2ϕ=1.
Therefore, c2=−c1, and c1(Φ+ϕ)=1, or c1=1/√5,c2=−1/√5. We can rewrite the solution as
Fn=Φn−(−ϕ)n√5
Since ϕn→0 as n→∞, we see that Fn→Φn/√5, and Fn+1/Fn→Φ